Choosing the two finalists
Economic Theory
ISSN 0938-2259
Volume 46
Number 2
Econ Theory (2010)
46:211-219
DOI 10.1007/
s00199-009-0516-3
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Econ Theory (2011) 46:211–219
DOI 10.1007/s00199-009-0516-3
RESEARCH ARTICLE
Choosing the two finalists
Kfir Eliaz · Michael Richter · Ariel Rubinstein
Received: 17 September 2009 / Accepted: 22 December 2009 / Published online: 5 January 2010
© Springer-Verlag 2010
Abstract This paper studies a decision maker who for each choice set selects a
subset of (at most) two alternatives. We axiomatize three types of procedures: (i) The
top two: the decision maker has in mind an ordering and chooses the two maximal
alternatives. (ii) The two extremes: the decision maker has in mind an ordering and
chooses the maximal and the minimal alternatives. (iii) The top and the top: the decision maker has in mind two orderings and he chooses the maximal element from
each.
Keywords Choice correspondence · Two-stage choice · Consideration sets ·
Axiomatization
JEL Classification
D01
1 Introduction
Choice theory aims to provide a framework in which observations about a decision
maker’s behavior may be organized conceptually. A decision maker is described by
We wish to thank the two referees for their valuable comments.
K. Eliaz (B)
Department of Economics, Brown University, Providence, USA
e-mail: [email protected]
M. Richter · A. Rubinstein
Department of Economics, New York University, New York, NY, USA
A. Rubinstein
University of Tel Aviv Cafés, Tel Aviv, Israel
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the possible choices he would make in all feasible situations that the decision maker
might confront within a certain context. These choices are represented by a choice
function or a choice correspondence. The typical interpretation of both is that the
decision-maker will ultimately “consume” only a single alternative from the available
set. Thus, if the choice correspondence selects more than one alternative, we interpret that to mean that the single element, which the decision-make will ultimately
consume, may be any of the selected elements (not that the selected elements form
a bundle, which the decision-maker will consume all at once). We have in mind a
situation where a decision maker is observed choosing small sets, out of any possible
subset of alternatives. Here, we focus on the case that the subsets consist of at most two
elements. We will comment on how our analysis would change if we were to assume
that the decision maker cannot choose more than some arbitrary fixed number of elements. We offer two interpretations for the decision-maker’s choice of more than one
element:
(i) Consideration sets. When the decision maker chooses a set in our framework he
actually selects a consideration set. That is, the decision maker employs a decision
process in which he first selects a small number of alternatives that draw his attention, or that he finds deserving a closer look at the second and final stage (see Eliaz
and Spiegler 2009). When the decision maker is an individual, this may reflect an
attempt to minimize the number of considered elements as deliberating on all of them
is too “costly”. When the decision maker is an organization, the choice procedure
may involve two distinct agents, one who narrows down the list of possible options
to a small set of contenders, and a top manager who decides on the final choice (see
García-Sanz 2008).
(ii) A team. The chosen set is not an input for another decision problem, but is itself
the final decision. This interpretation fits situations in which the decision maker must
choose a team with two members. Examples of such situations include the choice of
candidates for the position of president and vice president in an election, selecting a
two-pilot crew for an airplane, selecting a pair of tennis players for a doubles match, or
choosing a pair of instructors for teaching the microeconomics and macroeconomics
courses in a graduate program.
Formally, let X be a (finite) set of alternatives. A choice correspondence D is taken
in this paper to be a function which assigns to every non-empty A ⊆ X a non-empty
subset of A of size at most two.
With regards to our first interpretation, the framework we study here is in the spirit of
most of the decision-theoretic literature on preferences over menus (e.g., Kreps 1979)
which visualizes the decision process as a two stage process and axiomatizes only the
first stage which contains the selection of the second-stage choice problem, ignoring
the actual choice in the second stage. An alternative framework, would describe a
decision maker by a function that assigns to every choice set A, a subset B ⊆ A and
an element a ∈ B with the interpretations that B is the set of the finalists and a is the
final chosen element.
Our objective is to axiomatize the following three types of choice correspondences
(note that we use the term “ordering” to denote a strict total order):
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I. The top two. The decision maker has in mind an ordering and he chooses the two
best elements according to this ordering. Under the consideration set interpretation,
this procedure fits situations in which the decision maker proceeds in two steps. He
first narrows the set of options to only a few contenders by using some initial criterion
represented by an ordering, which does not necessarily coincide with his true preferences (for example, the simplest two alternatives). The elements chosen in the initial
stage (which constitute the decision-maker’s consideration set) are examined more
closely at a later stage, where the decision-maker then applies his true preferences.
Under the team selection interpretation, this procedure means that the two individuals
are chosen independently using the same preference criterion.
II. The two extremes. The decision maker orders the alternatives along one dimension
and chooses the two most extreme options. Under the consideration set interpretation,
this procedure fits situations where a decision maker thinks that his choice is actually
between two different “directions”. He therefore selects a consideration set consisting
of the two extremes with the aim of choosing the direction in the second stage. We
may also interpret the choice of the two extremes as the choice of two “attention-grabbers”. Under this interpretation, the options are ordered along some dimension (small
to big, cheap to expensive, ugly to handsome), and the consideration set consists of the
two extremes, the two elements that draw the most attention. Under the team selection
interpretation, the choice of the two extremes fits a desire for the most extreme variety.
III. The top-and-the-top. The decision maker has in mind two considerations. He
selects the (one or two) alternatives that are deemed the best by at least one of the
considerations. Of course, any two-extremes procedure may be thought of as a topand-the-top procedure where the two rationales are the ordering of the two-extremes
and its inverse. Note that if a top-and-the-top procedure assigns a couple to every
set, then each of its rationales must be the inverse of the other, and hence, it may be
described as a two-extremes procedure.
One may also interpret the different considerations in the top-and-the-top procedure
as the orderings of two distinct individuals (e.g., a married couple) who need to choose
from a set of options. The top-and-the-top selects the (at most) two extreme Pareto
efficient alternatives. Of course, the set of Pareto efficient alternatives may be bigger
(for a characterization of the Pareto efficient correspondence see Sprumont 2000).
Under the team interpretation, this procedure may be viewed as a way of preparing
for two possible scenarios that the decision maker may face: for each scenario, he
picks the best alternative to handle that contingency (if there is one alternative that is
best in both contingencies, he picks it).
Note that our paper differs from some recent papers, which have analyzed choice
functions that are consistent with two-stage choice procedures, in which the decision
maker first selects a subset of options to consider, and then chooses from that subset.
See, Manzini and Mariotti (2006), Manzini and Mariotti (2007), Masatlioglu et al.
(2008), and Manzini and Mariotti (2009). Contrary to us, in these papers, the final
choice is observed and the decision-maker’s “shortlist” or “attention-grabbers” are
inferred from the final choice. For a work on rationalization of a choice correspondence by any number of multiple rationales, see (Aizerman and Malishevski 1981).
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In the rest of this short paper we provide axiomatizations of the three choice correspondences discussed above.
2 Three axiomatizations
2.1 The top two
We will use three axioms in order to characterize the “top two” choice correspondence:
A0 : ∀A ⊆ X such that |A| ≥ 2, |D(A)| = 2.
A1: If a ∈ D(A) and a ∈ B ⊂ A then a ∈ D(B).
A1 is commonly referred to in the literature as Sen’s α Axiom. If an element is chosen
from a set, and it is also a member of a subset, then it is chosen from that subset as
well.
A2: For every set A and a ∈
/ D(A) = {x, y}, if a ∈ D(A\{x}) then a ∈ D(A\{y}).
A2 states that if an alternative is added to the set of selected alternatives when one of
the chosen elements is removed, then it is also selected when the other chosen element
is removed.
Proposition (The top two). A choice correspondence D satisfies A0, A1 and A2 iff
there exists an ordering on X such that D(A) consists of the two top elements in A
according to .
Proof It is trivial to verify that a top two choice correspondence satisfies the axioms
A0, A1, and A2.
Let D be a choice correspondence satisfying the three axioms. We will prove the
proposition by induction on the size of X . For the inductive step, let D(X ) = {x, y}
(well defined by A0).
By the inductive hypothesis, there exists an ordering on X − {y} such that D(A)
consists of the -top two elements for any A ⊆ X − {y}. By A1, x ∈ D(X − {y})
and thus, must be one of the two -top elements. Without loss of generality we can
assume x is at the top of . Extend on X − {y} to on X by putting y on top of
and letting be equal to otherwise.
We need to show that for every A that contains y, D(A) consists of the -top two
elements in A. By A1, y ∈ D(A) and indeed it is one of the -top elements in A. Let
z be the other element in D(A). If x ∈ A then by A1, z = x is the second top element
in A. If not, then by A2, z ∈ D(A − {y} ∪ {x}) (z was selected after x was removed
from A ∪ {x}, hence, it should also be selected when y is removed instead of x). By
the inductive step, z is the second -top element in A − {y} ∪ {x} and thus, also the
second -top element in this set. It follows that z is also the second -top element in
A (where y is replaced by x).
Note that the three axioms are independent.
The choice correspondence that selects only the top element clearly satisfies A1
and vacuously A2.
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The choice correspondence that selects the second and third elements from the top
according to some fixed ordering satisfies A0 and A2 but not A1.
The two extremes procedure satisfies A0 and A1 but not A2.
Comment: The above proposition could be extended as follows. Let D satisfy A1
and A2 and a modified version of A0, where D(A) is required to contain M elements
(unless A contains less than M elements, a case where D(A) is required to be equal
to A). Then there is an ordering on X such that whenever A contains more than
M elements, D(A) contains the top M elements in A. The proof (by induction on the
size of X ) is similar to the one of the above proposition and we provide it here for the
sake of completeness.
For the inductive step, let y ∈ D(X ). By the inductive hypothesis, there exists an
ordering on X − {y} such that D(A) consists of the -top M elements for any
A ⊆ X − {y}. By A1, the other M − 1 elements of D(X ) are in D(X − {y}) and thus,
must be one of the M -top elements in X . Without loss of generality we can assume
that they consist of the top M − 1 elements in . Extend on X − {y} to on X
by putting y on top of and letting be equal to otherwise.
We need to show that for every A that contains y, D(A) consists of the -top M
elements in A. Let z ∈ D(A). If z ∈ D(X ) then by A1, z is one of the M -top
elements in A. If not, then by A1, it must be that there is an element x ∈ D(X ) which
is not in A. By A2, z ∈ D(A − {y} ∪ {x}) (z was selected after x was removed from
A ∪ {x}, hence, it should also be selected when y is removed instead of x). By the
inductive step, z is one of the -M top elements in A − {y} ∪ {x} and thus, one of
the -M top elements in this set. It follows that z is also one of the M -top element
in A (where y is replaced by x).
2.2 The two extremes
To axiomatize the “two extremes” choice correspondence we use A0 and a new axiom
that states that if a is selected when the alternative x is added to a set A and also when
the alternative y is added to A, then a is selected after both x and y are added.
A3: ∀A ⊆ X such that |A| ≥ 2 and x, y ∈ A, if a ∈ D(A∪{x}} and a ∈ D(A∪{y}),
then a ∈ D(A ∪ {x, y}}.
Lemma A choice correspondence D that satisfies A0 and A3 also satisfies A1.
Proof Let D be a choice correspondence that satisfies A0 and A3 but not A1. Take
A to be a minimal subset of the ground set X such that D violates A1 on A. Denote
D(A) = {a, z}. The minimality of A means that there is some x = a such that
a ∈
/ D(A − x) (or similarly with z) and that A1 holds for all smaller subsets of A.
Notice that if for some b ∈ X , b ∈ D(A − x) and b ∈ D(A − y), then by A3,
b ∈ D(A) and hence, b ∈ {a, z}.
Case (i): a ∈
/ D(A − z). Let D(A − z) = {c, d} and D(A − a) = {e, f }. One of
the elements in {e, f }, say e, is not z. Now e cannot be c or d because
otherwise, it would be a member of D(A) though it is neither a nor z. By
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the fact that A1 holds for A − a and A − z the three distinct elements c, d, e
are all members of D(A − a − z), violating A0.
Case (ii): a ∈ D(A − z) and a ∈
/ D(A − c) for some c ∈
/ {a, z}. As before, c ∈
D(A − a) or c ∈ D(A − z). If c ∈ D(A − a), then since A1 holds for A − a
and A − c, we have D(A − c) = D(A − a − c) = D(A − a). But by A3 and
A0, D(A − a) = D(A), a contradiction. Suppose c ∈ D(A − z) = {a, d}.
As before, at least one of the elements of D(A − c), say e, is not z. Hence,
the three distinct elements, a, d and e, are all members of D(A − z − c),
violating A0.
Proposition (The two extremes): A choice correspondence D satisfies A0 and A3
iff there exists an ordering on X such that D(A) consists of the -maximal and
-minimal alternatives in A.
Proof Any two-extremes choice correspondence satisfies A0. To see that it satisfies
A3, let A be a set such that |A| ≥ 2 and let a be a member of both D(A ∪ {x}) and
D(A ∪ {y}). WLOG let a = max(A ∪ {x}, ). Then a = max(A, ). It follows that
a = min(A∪{y}, ), hence a = max(A∪{y}, ). Therefore, a = max(A∪{x, y}, )
and thus, a ∈ D(A ∪ {x, y}).
Let D be a choice correspondence that satisfies A0 and A3. By the lemma, D also
satisfies A1. Let D(X ) = {L , R}.
For any two distinct elements a and b, we say that a “is to the left of” b, and denote
a → b, if D({a, b, R}) = {a, R} and D({L , a, b}) = {L , b}. By definition, → is an
asymmetric relation.
We first verify that the relation “to the left of” is total. Notice that by A1, L → a
for any a = L and a → R for any a = R. Consider a, b distinct from L , R.
By A1, R ∈ D({a, b, R}) and L ∈ D({L , a, b}). If D({a, b, R}) = {a, R}, then
D({L , a, b}) = {b, L} (and a → b), because if D({L , a, b}) = {L , a}, then by A3,
a ∈ D({L , a, b, R}) but by A1, D({L , a, b, R}) = {L , R}.
We now show that the relation “to the left of” is also transitive. Assume a → b
and b → c. That is, D({a, b, R}) = {a, R}, D({L , a, b}) = {L , b}, D({b, c, R}) =
{b, R} and D({L , b, c}) = {L , c}. By A0 and A1, D({a, b, c, R}) = {a, R} and by
A1, D({a, c, R}) = {a, R}. Similarly, D({L , a, b, c}) = {L , c} and D({L , a, c}) =
{L , c}. Hence, a → c.
Finally, let A ⊆ X and denote by l and r the maximal and the minimal element of
the “to the left of” relation. We will show that D(A) = {l, r }. Assume D(A) does not
contain l. The set D(A ∪ {R}) contains R (by A1 and D(X ) = {L , R}) but not l (since
if it were, then by A1, it would also be included in D(A)). Thus, D(A ∪ {R}) = {a, R}
for some a = l. By A1, D({l, a, R}) = {a, R} contradicting the definition of l as being
to the left of a. Similarly, D(A) contains r .
The two axioms used above are independent.
The choice correspondence that chooses the single top element satisfies A3 and A1
but not A0.
The top-two choice correspondence satisfies A0 and A1 but not A3 (if the third
best element in X is the top in A, then it is selected when either the best or second
best element in X is added, but not when both are added).
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2.3 The top and the top
In order to characterize the top-and-the-top procedure, we use A1, a variation of A0
and two additional axioms.
A0’: ∀A ⊆ X, |D(A)| ≤ 2.
A0’ modifies A0 to allow for the possibility that the two rationales agree on the
maximal element on some choice set.
A4: If D(A) = {a} and a ∈ B ⊆ A, then D(B) = D(A).
A4 requires that if D(A) is a singleton, which is also a member of a subset B, then
D(B) is this singleton. It is a weakening of the traditional Independence of Irrelevant
Alternatives, D(A) ⊆ B ⊆ A ⇒ D(B) = D(A). Note that A1 would guarantee that
a is a member of D(B) but not that D(B) = {a}. A1 and A4 imply the traditional
Independence of Irrelevant Alternatives in the presence of A0’ (or A0).
A5: If a is a member of each of the sets, D(A1 ), D(A2 ) and D(A1 ∩ A2 ), and also
D(A1 ∩ A2 ) contains two elements, then a ∈ D(A1 ∪ A2 ).
A5 captures situations where the decision maker is choosing at most two alternatives from every set using two “reasons” R1 and R2 . Interpret a Ri b to mean that Ri is
a reason for choosing a and not b. An element a is chosen from a set A if there is a
reason i for which a Ri x for all x in A. A decision maker who follows this procedure
satisfies the following: if both a and b are chosen from A1 ∩ A2 , then it must be that
for one of the two reasons, say R2 , b R2 a. Thus, it must be that a is chosen from A1
and A2 by applying the same reason R1 . Thus, a R1 x for any x in either A1 or in A2 ,
and thus, it is chosen also from A1 ∪ A2 .
Proposition (The top and the top) A choice correspondence D satisfies Axioms A0’,
A1, A4, and A5 iff there are orderings 1 , 2 (possibly identical) such that D(A) =
{max(A, 1 ), max(A, 2 )}.
Proof The top and the top procedure clearly satisfies A0’, A1 and A4. As to A5,
assume x ∈ D(A1 ), D(A2 ) and D(A1 ∩ A2 ) = {x, y}, where y = x. The element y
maximizes one of the orderings, say 1 , in A1 ∩ A2 so that y 1 x. Thus, x must be
the 2 -maximal element in both A1 and A2 and therefore in A1 ∪ A2 , which implies
x ∈ D(A1 ∪ A2 ).
Let n be the number of elements in X and let D be a choice correspondence, satisfying the four axioms. We construct inductively two rankings a1 1 · · · 1 an , and,
b1 2 · · · 2 bn and then show that D(A) = {max(A, 1 ), max(A, 2 )}. For ease
of notation, for i ≤ j let us denote Ai, j ≡ {ai , . . . , a j } and Bi, j ≡ {bi , . . . , b j }. For
convenience, denote A1,0 ≡ B1,0 ≡ ∅.
Assume that we have deduced the first m ≥ 0 elements of both orderings, then the
algorithm finds the m + 1st element of each ordering by considering the following
three cases:
Case 1: |D(X \A1,m )| = 1. Then we assign am+1 to be the element chosen by D
from X \A1,m .
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Case 2: |D(X \A1,m )| = 2 and B1,m = A1,m . Then we set am+1 , bm+1 to be the two
members of D(X \A1,m ).
Case 3: |D(X \A1,m )| = 2 and B1,m = A1,m . Then there is an element bi ∈ X \A1,m
where i ≤ m. Let bi ∗ be the element of X \A1,m with the minimal index.
By construction, bi ∗ ∈ D(X \B1,i ∗ −1 ) and X \A1,m ⊂ X \B1,i ∗ −1 . By A1,
bi ∗ ∈ D(X \A1,m ). Define am+1 to be the other element in D(X \A1,m ).
To determine bm+1 , apply the above algorithm to D(X \B1,m ). Note that when
|D(X \A1,m )| = 2 and B1,m = A1,m , then D(X \B1,m ) = D(X \A1,m ), and applying the above algorithm to either D(X \A1,m ) or D(X \B1,m ) yields the same pair,
(am+1 , bm+1 ).
Consider an arbitrary set A. We have constructed 1 , 2 and need to show that
D(A) = {max(A, 1 ), max(A, 2 )}. Let ai = max(A, 1 ) and b j = max(A, 2 ).
Note that A ⊆ Ai,n and A ⊆ B j,n , and hence, ai , b j ∈ D(A) by A1. If ai = b j , then
by A0 we have D(A) = {ai , b j }.
It remains to verify that D(A) = {x} whenever ai = b j = x. If |D(Ai,n )| = 1 (or
if |D(B j,n )| = 1), then by A4 and the fact that A ⊆ Ai,n (respectively, A ⊆ B j,n ),
we have that D(A) = D(Ai,n ) = {x}. Suppose |D(Ai,n )| = |D(B j,n )| = 2. Denote
C∩ ≡ Ai,n ∩ B j,n and C∪ ≡ Ai,n ∪ B j,n . Since A ⊆ C∩ , then by A4 all we need to verify is that D(C∩ ) = {x}. By construction, x = ai ∈ D(Ai,n ) and x = b j ∈ D(B j,n ),
thus by A5, it is sufficient to show that x ∈
/ D(C∪ ). We achieve this by proving that
D(C∪ ) contains two elements that are distinct from x. This would imply that one of
the conditions of A5 is violated, and since x ∈ D(Ai,n ), D(B j,n ), it must be that
D(C∩ ) = {x}.
From our construction of the two orderings, 1 and 2 , and from our assumption that |D(Ai,n )| = 2, it follows that ai = max(Ai,n , 2 ) ≡ bl and similarly,
b j = max(B j,n , 1 ) ≡ ak . Moreover, x, ak , and bl are all distinct because ak 1
b j = x = ai 1 bl . In addition, note that by the definition of ak , and by the observation that ak 1 ai , it follows that Ai,n ⊂ Ak,n and also B j,n ⊂ Ak,n . Similarly,
B j,n ⊂ Bl,n and also Ai,n ⊂ Bl,n . Therefore, C∪ is contained in both Ak,n and Bl,n .
By construction, ak ∈ D(Ak,n ) and bl ∈ D(Bl,n ). Hence, by A1 and A0 , D(C∪ )
must contain both ak and bl , which are distinct from x.
The four axioms used for the proposition above are independent:
(i). The correspondence D(A) ≡ A satisfies all axioms except A0 .
(ii). Consider the following variant of the top-and-the-top procedure. Let 1 , 2 be
a pair of linear orderings on X . For all A ⊆ X with |A| ≥ 3, let
D(A) = {max(A, 1 ), max(A, 2 )}
and for all pairs {a, b} ⊂ X, let
D({a, b}) = {max({a, b}, 1 )}
This satisfies all axioms except A1.
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(iii). Let be a linear ordering on X. For all A with three or more elements, let
D(A) = {max(A, )}, while for all pairs A, let D(A) = A. This satisfies all
axioms except A4. To verify that A5 is satisfied, note that D(A ∩ B) contains
two elements only if A ∩ B is a pair. If A ⊆ B (or B ⊆ A), then A5 is vacuous. Otherwise, we have that |A|, |B| ≥ 3 and A5 is again vacuously satisfied
because {a} ∈ D(A), D(B) implies a = max(A, ) = max(B, ) and then
a = max(A ∪ B, ) which implies D(A ∪ B) = {a}.
(iv). Let D be the top-two procedure. This procedure satisfies A0 , A1, and A4
(vacuously). To see that A5 is violated when |X | ≥ 4 , denote the top four elements of X by a1 , a2 , a3 , a4 . Then, a3 ∈ D(X \a2 ) ∩ D(X \a1 ) and {a3 , a4 } =
D(X \a1 , a2 ). Therefore, we should have that a3 ∈ D(X \a1 ∪ X \a2 ) = D(X ),
but this is false.
Comment: An open question remains of how this procedure may be extended to a
procedure that selects the top of M > 2 orderings.
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